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As RD Sharma Class 12th Exercise 14.1 material contains solved questions, students can refer to them and quickly complete their portion.

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RD Sharma materials have very high standards when it comes to subject information. It gives students a clear insight into the subject and helps them better understand the concepts. Check out RD Sharma Class 12 Solutions Mean Value Theorem Ex 14.1, for solutions from RD Sharma’s book.

What are mean and median?

The average sum of all values is called the mean, whereas the median is the middle value from the given list of values. To learn about Mean Value Theorem, check RD Sharma Class 12 Solutions Chapter 12 Ex 14.1.

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RD Sharma Class 12 Exercise 14.1 Mean Value Theorems Solutions Maths

Updated on 27 Jan 2022, 03:29 PM IST

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RD Sharma Class 12 Solutions Chapter 14 Mean Value Theorems - Other Exercise

Mean value theorem exercise 14.1 question 1(i)

Answer:
Rolle’s Theorem is not applicable
Hint:
$f (x)$is continuous for al lx and hence continuous in$[1,3]$.
Given:
$f (x) = 3 + ( x -2 ) ^ \frac{2}{3}$on$[1,3]$
Explanation:
We have
$f (x) = 3 + ( x -2 ) ^ \frac{2}{3}$ …(i)

Being polynomial$f (x)$is continuous for all x and hence continuous in$[1,3]$.

2.
$\\f{}' (x) = \frac{2}{3} ( x -2) ^{\frac{2}{3}-1} \\\\ = \frac{2}{3} (x-2 ) ^{\frac{-1}{3}}$
Which exists in $(1,3)$
$f (x)$is derivable in$(1,3)$
3.
$\\f (1) = 3 + ( 1-2) ^{\frac{2}{3}} = 3 + (-1)\frac{2}{3}\\\\ = 3-1 = 2 \\\\ f (3) = 3 + ( 3-2)^\frac{2}{3} = 3 + ( 1) ^\frac{2}{3} \\\\ = 3+1 = 4 \\\\ f (1) \neq f (3 )$
Thus, third condition of Rolle’s Theorem is not satisfied.
Hence, Rolle’s Theorem is not applicable.

Mean value theorem exercise 14.1 question 1(ii)

Answer:

Rolle’s Theorem is not applicable.
Hint:
Rolle’s Theorem is applicable to function$f : [a,b] \rightarrow R$the following three conditions of Rolle’s Theorem is satisfied.
Given:
$f (x) = [x]$ for $x \in [-1,1]$
Explanation:
We have
$f (x) = [x]$,$x \in [-1,1]$
The greatest integer function $f$ is neither continuous in close interval$[-1,1]$, nor differentiable in open interval$(-1,1)$.
$f (-1)= [-1] = -1$ and $f (1)= [1] = 1$
$\therefore f (-1) \neq f (1)$
Hence, the Rolle’s Theorem is not applicable to $f (x)= [1] \ x \in (-1,1)$

Mean value theorem exercise 14.1 question 1(iii)

Answer:
Rolle’s Theorem is not applicable on$f (x)$in$[-1,1]$
Hint:
Let $\lim _{h \rightarrow 0 } \sin( \frac{1}{h} ) = k$as$k \in [-1,1]$
Given:
$f (x) = \sin \frac{1}{x} , x \in [-1,1]$
Explanation:
$\begin{aligned} L . H . S &=\lim _{x \rightarrow(0-h)} \sin \left(\frac{1}{x}\right) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{1}{0-h}\right) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{-1}{h}\right) \\ &=-\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \\ & = -k \ \ \ \ \ \ \ \ \ \left [ \because \lim _{h \rightarrow 0} \sin \frac{1}{h}= k,k \in [-1,1] \right ]\\\\ R . H . S &=\lim _{x \rightarrow(0+h)} \sin \left(\frac{1}{x}\right) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \ \ \ \ \ \ \\ &=k \\ \Rightarrow \quad & L \cdot H . S \neq R \cdot H . S \end{aligned}$

$f(x)$is not continuous at 0.
So, Rolle’s Theorem is not applicable on $f(x)$in$[-1,1]$

Mean value theorem exercise 14.1 question 1(iv)

Answer:
Rolle’s Theorem is not applicable [1,3]
Hint:
$f (x)$ is continuous for all x and hence continuous in [1,3]
Given:
$f (x) = 2x^2 - 5x +3$ on [1,3]
Explanation:
We have
$f (x) = 2x^2 - 5x +3$
1.Being polynomial$f (x)$ is continuous for all x and hence continuous in [1,3]
2. $f (x') = 4x -5$, which exists in (1,3)
$\therefore f (x)$ is derivable in (1,3)
3.
$\begin{array}{l} \\ f(1)=2(1)^{2}-5(1)+3\\ \begin{aligned} &=2-5+3=-3+3=0 \\ f(3)=& 2(3)^{2}-5(3)+3 \\ &=2 \times 9-15+3 \\ &=18-15+3=3+3=6 \\ \therefore f(1) & \neq f(3) \end{aligned} \end{array}$
Thus third condition of Rolle’s Theorem is not satisfied.
Hence, Rolle’s Theorem is not applicable.

Mean value theorem exercise 14.1 question 1(v)

Answer:
Rolle’s Theorem is applicable [-1,1]
Hint:
f(x) is continuous for all x and hence continuous in [-1,1]
Given:
$f (x) = x ^{\frac{2}{3}}$ on [-1,1]
Explanation:
We have
$f (x) = x ^{\frac{2}{3}}$

Being polynomial f(x) is continuous for all x and hence continuous in [-1,1]
$f ' (x) = \frac{2}{3} x ^{\frac{-1}{3}}$ , which exists in (-1,1)

$\therefore f (x)$is derivable in (-1,1)
3.
$\\f (-1) = (-1)^{\frac{2}{3} } = 1 \\\\ f (1) = (1) ^{\frac{2}{3}} = 1 \\\\ f (-1) \neq f (1)$

Thus third condition of Rolle’s Theorem is satisfied.

Hence, Rolle’s Theorem is applicable.

RD Sharma Class 12th Exercise 14.1 contains the chapter Mean Value Theorem. It has a total of 40 questions that are of Level 1 difficulty. It contains basic concepts like calculating the mean of functions for the given parameters, Rolle's theorem, etc. As this is an introductory exercise, these sums are relatively fundamental and easy to answer. But due to the high number of questions, students get panicked and spend most of the time solving them.

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